Each wheel of a motorcycle is of 600 mm diameter and has moment of inertia of 1.2 kg-m2. The total mass of the motorcycle and the rider is 180 kg and combined center of mass 580 mm above the ground level when the cycle is upright. The moment of inertia of the rotating parts of the engine of heel necessary when the motor cycle takes a turn of 35m radius at speed of 54 km/h. -

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: Dec 2015

$dw= 600 mm = 0.6 m –– rw= 0.6/2= 0.3 \\ Iw= 1.2 kg-m^2 \\ M= 200 kg \\ n = 580 mm= 0.58m \\ I_E= 0.2 kg-m2 \\ G= 5 = W_E/W_W \\ R= 35 m$

$V= 54 km/n= 54 \times 50/180= 15 m/sec.$

To find the angle of heel for negotiating a curve,

We knew that velocity of the vehicle is-

$V= rw \times ww \\ \therefore 1.5= 0.3 \times ww \\ Ww = 50 rad/sec.$

But the gear ratio is 5

$G= W_E/W_W = 5 = W_E/50= W_E=250 rad/sec.$

WE = 250 rad/sec.

Velocity is also given by

$V= R \times Wp \\ 15= 35 \times Wp \\ Wp= 15/35= 0.4285 rad/sec.$

Angle of heel is given by

$tanѳ= \dfrac{[MR Wp2×h+(2 Iw×Ww– IE×WE) Wp]}{mgh}$

$tanѳ$ $=\dfrac{200 \times 35 \times (0.4285)2 \times 0.58+(2 \times 1.2 \times 50+0.2 \times 250) \times 0.4285}{200 \times 9.81 \times 0.58} \\ = \dfrac{745.46+72.845}{200 \times 9.81 \times 0.58} = \dfrac{818.305}{200 \times 9.81 \times 0.58}$

$\tan ѳ= 0.7190 \\ Ѳ=35.719°$